Re: Tracking

Wayne Throop wrote:
: "Logan Kearsley" <chronosurfer@xxxxxxxxx>
: I shall attempt to explain. If 'normal' (by which I mean, having a real
: wavelength) radiation is emitted from an object moving FTL, it could
: behave as though the frame of reference of the object is just like any
: other inertial frame. In a non-FTL frame, then, the emitted radiation
: would seem to have imaginary wavelength, whatever that means, unless it
: somehow works out that velocities add up so that radiation emitted to
: the aft is no longer moving FTL.

I still don't follow. Velocities of photons don't add to those
of the emitting object, so ... well, I still don't follow.

Well, radiation doesn't just have to mean photons. But I was mostly
thinking of luxons, so reading on...

: Or, the radiation could behave as though it was emitted from an object
: in the same place as the FTL object, but moving with non-FTL velocity.

But the velocity of the emitting object doesn't matter... ah, I think
I see. Doesnt' matter wrt velocity; does matter wrt other things.

Unless it's particulate radiation, in which case it does matter wrt
velocity. I'm trying to talk about all cases at once and getting
confused.... Keeping just to the case of photons, though, yes, doesn't
matter wrt velocity, does matter wrt other things (like wavelength).

So you say "unless" it somehow works out to apply to velocity, then
it must apply to the wavelength. But we know that that's not how it
works out; the velocitiy of emitted photons doesn't add. Therefore,
relativistic doppler applies to the wavelength, which becomes imaginary.
The wavelength of forward-emitted photons from an object moving at c is
zero (infinite energy), and for rearwards-emitted photons is infinite
(zero energy). So going beyond that, things get all imaginary-like.
Um... if there were proper time to emit anything at all anyways,
which there isn't.

So, while it still seems to me to be a category error to say "emit
radiation in a frame", I think I see what you're getting at. I think.

I think you think correctly. I think.

Um. Hey, maybe the proper time being imaginary, multiplied by the
wavelength being imaginary, means the emitted photon would be OK. Heh.

Oo! But wait... wouldn't that allow some emitted photons to have
negative energy?

-l.

.

Relevant Pages

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